uva 202求循环小数

 The decimal expansion of the fraction 1/33 is , wherethe is used to indicate that the cycle 03 repeatsindefinitely with no intervening digits. In fact, the decimal expansion ofevery rational number (fraction) has arepeating cycle as opposed to decimal expansions of irrational numbers,which have no such repeating cycles.

Examples of decimal expansions of rational numbers and their repeating cyclesare shown below. Here, we useparentheses to enclose the repeating cycle rather than place a bar overthe cycle.

Write a program that reads numerators and denominators of fractions anddetermines their repeating cycles.

Forthe purposes of this problem, define a repeating cycle of a fraction to bethe first minimal length string of digits tothe right of the decimal that repeats indefinitely with no intervening digits.Thus for example, the repeating cycle ofthe fraction 1/250 is 0, which begins at position 4 (as opposed to 0 whichbegins at positions 1 or 2 and as opposedto 00 which begins at positions 1 or 4).

Input

Each line of the input file consists of an integer numerator, which isnonnegative, followed by an integerdenominator, which is positive. None of the input integers exceeds 3000.End-of-file indicates the end of input.

Output

For each line of input, print the fraction, its decimal expansion through thefirst occurrence of the cycle to the rightof the decimal or 50 decimal places (whichever comes first), and the lengthof the entire repeating cycle.

In writingthe decimal expansion, enclose the repeating cycle in parentheses when possible.If the entire repeating cycle doesnot occur within the first 50 places, place a left parenthesis where thecycle begins - it will begin within the first 50places - and place ``...)" after the 50th digit.

Print a blank line after every test case.

Sample Input

76 255 431 397

Sample Output

76/25 = 3.04(0)   1 = number of digits in repeating cycle5/43 = 0.(116279069767441860465)   21 = number of digits in repeating cycle1/397 = 0.(00251889168765743073047858942065491183879093198992...)   99 = number of digits in repeating cycle

循环小数，仔细就行。下面代码主干是对的，不过最后的输出格式被我后来存到自己u盘里时改了-.-改回来其实挺容易的：

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int main(){     int a,b,i,k;     while(cin>>a>>b)     {            int flag=1,a0=a;            int yu[3000],shang[3000];         for(i=0;1;i++)         {            shang[i]=a/b;             yu[i]=a-shang[i]*b;             if(yu[i]==0)             {                 k=i;                 i++;                 shang[i]=0;                 yu[i]=0;                 break;             }            for(k=0;k<i;k++)            if(yu[k]==yu[i])            {                flag=0;                break;            }            if(flag==0)break;            a=yu[i]*10;         }         cout<<a0<<'/'<<b<<" = "<<shang[0]<<'.';            for(int x=1;x<=k;x++)            cout<<shang[x];            cout<<'(';int x;         for(x=k+1;x<=i;x++)            cout<<shang[x];            cout<<')'<<endl;         cout<<"循环节长度是"<<i-k<<endl<<endl;     }    return 0;}